Question: Each year, Francesca earns a salary that is $2\%$ higher than her previous year's salary. In her first $5$ years at this job, she earned a total of $\$187{,}345$. What was Francesca's salary in her $1^{\text{st}}$ year at this job? Round your final answer to the nearest thousand.
Explanation: Notice that Francesca's salaries over the years form a geometric sequence. The total amount Francesca earns after $ n$ years is the ${\text{sum}}$ of the first $n$ terms in the sequence. This is called a geometric series. This is the formula for that sum: $ S={a}\left(\dfrac{1-{r}^{ n}}{1-{r}}\right)$ where ${a}$ is the first term and ${r}$ is the common ratio. We can use this formula, along with the given information, to find Francesca's salary in her $1^{\text{st}}$ year at this job, $ a$. Using the given information We are given that Francesca's salary each year is ${2\% \text{ higher}}$ than it was in the previous year. So we'll use a common ratio of ${1.02}$ for $ r$. We are given that she earned a total of ${\$187{,}345}$ in her first $ 5$ years. So the sum of series $ S$ is ${\$187{,}345}$, and the number of terms $ n$ is $ {5}$. We are looking for her salary in her $1^{\text{st}}$ year at this job, $ a$. Finding the first term $\begin{aligned} {187{,}345}&={a} \cdot \dfrac{1-\left({1.02}\right)^{{5}}}{1-\left({1.02}\right)} \\\\ \dfrac{1-\left({1.02}\right)}{1-\left({1.02}\right)^{{5}}} \cdot {187{,}345} &= {a} \\\\ 35{,}999.91 &= {a} \end{aligned}$ Answer To the nearest thousand, Francesca's salary in her $1^{\text{st}}$ year at this job was $36{,}000$ dollars.